# 1998 Grand Caravan

Page 4 of 13
• posted on March 25, 2006, 12:25 am

ROTFLMBO!!!!! Gawd you are sure full of funny statements.
If a ratio results in a reduced ability to perform an operation or function for a given input, the ratio is less efficient.
Lever anaology: the longer the handle side of a lever, the more efficiently a person can do a job for a given input force.

And changing final drive ratios and adding underdriven transmission ratios results in greaterr efficiency.

For someone that keeps putting me down you have very little grasp of reality. Obviously you are set in the ways you were brainwashed into.
--
Budd Cochran

John 3:16-17, Ephesians 2:8-9
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• posted on March 25, 2006, 1:05 am
Budd Cochran wrote:

Yes, I'm sure correct statements are pretty funny to someone without a clue about physics.

No, it isn't more efficient at all as you have to move the lever farther to accomplish the same amount of work on the other end. The longer the handle the less force is required, but more distance is required in direct proportion to the reduced force. Since work is a force acting through a distance, the end result is no change in efficiency at all.
Since you aren't able to show the math or physics as you several times claimed you could, here's a really simple example using a lever which you claim to understand. I won't complicate things with gears, but they are just two levers touching each other, but that adds a little complexity and you don't deal well with that.
Definition: Efficiency = work out/work in (the / means "divided by")
Here's a reference simple enough that you should be able to understand it: http://www.s-cool.co.uk/topic_quicklearn.asp?loc=ql&topic_id4&quicklearn_id=1&subject_idh&ebtC5&ebn=&ebs=&ebl=&elc 
Definition: Work = force * distance (the * means "multiplied by")
Again, here's a pretty simple reference (high school level): http://www.glenbrook.k12.il.us/gbssci/phys/CLass/energy/u5l1a.html
Now we have a lever that is 10 feet long with the fulcrum in the middle. Let's say we put a 100 lb. weight on one end. If we move the other end of the lever down by 1 ft we will lift the weight 1 ft. The work out is the work done on the weight, which is 100 ft-lbs. The work in is the work done by whatever pushed down on the lever and this is also 100 ft-lbs since we had a force of 100 lbs acting through a distance of 1 ft. Thus the efficiency is 100/100 or 1.
Now shift the fulcrum such that it is 9 feet from the weight and 1 foot from whatever or whoever is pushing on the lever. Now to lift the 100 lb weight, we will have to push down with a force of 900 lb. You keep claiming that we have now lost efficiency since it now requires more force on the lever than before. However, let's do the math. Again, let's raise the weight 1 ft to keep the work out the same as before at 100 ft-lbs. How far does the lever have to be pushed down to raise the weight 1 ft? Since the ratio is 1:9, we have to push the lever down only 1/9th of a foot to raise the weight one foot. So how much work does this require? It requires force * distance or 900 lbs * 1/9 ft which equals ... drum roll please ... 100 ft-lbs. Now lets calculate our efficiency again. It is equal to 100/100 or 1. Doesn't this sound familiar? Let's see, why yes, it is the SAME as when the ratio of the lever was 1:1. Imagine that, the lever ratio doesn't change the efficiency at all. :-)

If you equate understanding basic physics and math with being brainwashed, then I'm guilty as charged. OK, I demonstrated pretty clearly above that changing the ratio of a lever has NO affect on the efficiency of the lever. It changes the ratio of forces, but has ZERO affect on the efficiency of work/power transfer.
Now it is your turn to show a clear example that supports your assertion that changing a ratio does change the efficency.
Matt
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• posted on March 25, 2006, 1:43 am
Budd Cochran wrote:
<snip>

<snip>
Hi...
Here's the answer for you all. You're simply using the word "efficient" differently.
To the engineer it's simply a mathematical formula. ie; 12 volts in and 11 volts out equals blah blah blah.
To the mechanic; it's more practical. ie; if you can't reach the ignition key to crank the starter, then it's blah blah blah
There it is, now all of you sit down, have a beer or whatever, and get back to helping others and each other fix your cars efficiently :)
Ken
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• posted on March 25, 2006, 1:54 am
Ken Weitzel wrote:

Exactly, that is what we've been trying to show Budd. He claimed his argument was based on physics, so I used the physics definition of efficiency. He then started making up his own definitions, but continued to claim they were based on "physics."
Maybe his definition of physics is also other-worldly. :-)
Matt
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• posted on March 24, 2006, 12:52 am
Have you ever heard of a guy by the name of Archimedes? Look up his information on levers.
The same rules apply to gearing. An underdrive ration is like a prybar with a long handle and a short span from pivot to load, a small force moves a large mass. OTOH, an overdrive is like a short handle and a longer span from pivot to load and requires much more force to move a small mass.
To make a OD work, the axle ratio has to be lowered and it doesn't quite make up the losses in the drive train . . .it can't as that would violate the laws of levers and physics.
"Give me a lever long enough and a place to stand and I shall move the World." (Archimedes, if I remember the quote correctly)
--
Budd Cochran

John 3:16-17, Ephesians 2:8-9
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• posted on March 24, 2006, 2:18 am
Budd Cochran wrote:

Your argument isn't very convincing in that, in your lever examples, you appear to be equating force and mechanical advantage with energy loss, and that just isn't inherently the case (the same being true with gear ratios, which I know you correctly related to levers). The energy loss (heat generated) has nothing to do with the mechanical advantage (the distance/force trade-offs) but has everything to do with the friction in the transfer of motion - nothing to do with the lengths of the levers (size of the gears), and everything to do with the reduction of friction at the points of sliding, pivoting, etc.
No offense, but this is exactly the kind of technically incorrect discussion that seems to crop up with those who consider themselves technically savvy in a *hands-on* sense but who put down those with technical (in a *theoretical* sense) expertise. In the same way that you would correctly invoke Archimedes, I would say "Yes, but you also need to consider the law of the conservation of energy which you seem to be ignoring (in your lever examples)". You could move the world with Archimede's lever, but not very rapidly. You could use an opposite lever to move a pea and move it very rapidly. The energy put into the system to accomplish both may be the same - one has a lot of mass moving very slowly, the other has a tiny mass moving very rapidly - the energy calculations are the same - energy applied minus friction losses energy out.
Now - one implementation of the gears or levers may be more efficient than the other, but it's all in how the friction losses are managed, and not in the mechanical advantage per-se.
There's a place for both "hands on" and "theroetical" - and results are best when they co-exist. Real world follows the laws of physics - what some here disdainfully call "theoretical"; but what many are sarcastically or disdainfully calling the "theoretical" *is/are* the real world laws of physics.
Bill Putney (To reply by e-mail, replace the last letter of the alphabet in my address with the letter 'x')
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• posted on March 24, 2006, 5:35 am
Bill Putney wrote:

Transmissions 101
Frictional losses are a given in ANY transmission design, as are heat losses and parasitic losses. So, why mention the obvious that, as an engineer, you should know about already?
In an overdrive ratio, do you have more or less usable torque on the output shaft of the gearset? Energy ( torque in), divided ( OD, remember?) by the ratio equals output torque ( minus parasitic losses, of course!!!!) proportionally reduced according to the leverage . . er, ratio of the gearset.

Sorry, it seems I stepped on a toe or two . . .

You assume they are ignored because they are unmetioned. Name a transmission design that generates no friction, heat or parasitic losses????? Now, since there is no such thing, these losses are common and may be set asiode for a discussion of RATIOS AND THEIR EFFICIENCY.

True, I don't deny that and I didn't deny it. It was a metaphor.

Yaes, as in a trebuchet, for example ... nope, I didn't deny this either, bith both of your exaples are fine evdence of the effectiveness of RATIOS dependent, of course, on whether you want to lift a locomotive or throw rocks.
Gear ratios parallel that law.

Agreed, but why are you adding given and common known factors into a discussion on the effectiveness of RATIOS if not to cloud the topic to save your butt?

ROTFLMBO!!!!!!!!!!!
Assume a 20% loss in either transmission to parasitical loss for sake of discussion:
A 1:1 direct drive with a 100 ft.lb. input torque has 80 ft.lb.available from the output shaft out of a theoretical maximum of 100 ft.lb.
A 1:2 overdrive with a 100 ft.lb. input torque has only 40 ft.lb. left out of a theoretical maximum of 50 ft.lb. (This is still theoretical because it doesn't allow for the increased torque loss due to poor mechanical advantage)
Now, tell us again which has the best torque output to run 75 mph with?

Math is "real life", Bill. 1+1=2 And as long as math is a true science it can prove physics.
Budd
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• posted on March 24, 2006, 11:42 am
Budd Cochran wrote:

Budd - I'll say this in the least offensive way I know how (Matt said it and I said it before): Torque and energy are different units - losing torque thru levering or gearing is not in itself energy loss. Energy is work x speed x time. In your lever and gear examples, what you lose in torque you gain in speed - the product of the two is proportional to energy and will be the same - except - yes - for some energy loss due to friction, but that friction is not inherent in the lever or gear multiplication or division factor. Think about this example: Situation 1 - The fulcrum point is on a sliding friction bushing. Situation 2 - The fulcrum point is on a roller bearing of some sort. If you truly think it thru, you will realize that there are friction losses in both cases, but it is not directly related to the ratios. You will also have sliding friction between the gear teeth - but its amount is not a given relation to the torque multiplication factor - it can be varied all over the place depending on wheter it is lubed or not, and if lubed, what lube is used.
I will give you that, simplisitically speaking, if you have two meshed gears for one multiplication factor, you will have half the friction losses than if you have a series of two pairs of meshed gears to achieve the same final ratio (the difference between the OD and direct drive scenarios). But *that's* where the added losses are coming from that you're talking about - not in the ratios themselves.
In a nutshell, as I and Matt have already pointed out, your fallacy in your explanation is in trying to equate torque multiplication with energy multiplication. You trade off torque and speed, and energy being the product of the two (with time factored in), well - you do the math. The same kind of arguments occur about torque and horsepower because of the same failure on people's part to understand what they are mathematically (and in reality).
Bill Putney (To reply by e-mail, replace the last letter of the alphabet in my address with the letter 'x')
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• posted on March 24, 2006, 1:57 pm

Hold it!!! Hold the presses!! You just contradicted yourself. First you say there is no energy loss then there is . . you don't know squat.

What the fixation with friction losses?? Is it some thrill for you? How many times must I agree that there is parasitic losses before you set them aside and discuss ratios?

Reading comprhension isn't your strong point is it? Again, how many times will I hacve to tell you that I agree with this BEFORE IT GETS THRU YOUR THICK SKULL ( no, I am not a politically correct person, thank you)

I did the math and the net effect is a major measurable loss of efficiency.

No, the same arguments happen because overeducated idiots like yourself have the misconception you know everything.
The only reason you think I'm in error is because you added parasitic losses into a discussion ABOUT RATIOS.
So, you failed to discuss the topic.
--
Budd Cochran

John 3:16-17, Ephesians 2:8-9
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• posted on March 24, 2006, 10:18 pm
Budd Cochran wrote:

Show us the math then. First though, you need to look up efficiency and post that definition and then work from there.

No, we're more than happy to have you post some data, calculations, references to others who have done the calculations, etc. The fact is that you can't.
Matt
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• posted on March 25, 2006, 12:30 am

No, I don't have to post anything. Efficiency: the ratio of input to reslutant output. If your output isn't within the design parameter, it's either more or less efficient than required. Overdrive ratios are inheriently inefficient. the proof is the almost mandatory need to shift down to a lower ratio to climb a hill UNLESS YOU'RE TRAVELING AT A SPEED FAR ABOVE THE POSTED SPEED LIMIT.

The true fact is it wouldn't matter if I did. You'd just wave your wallpaper around and find some fault with it, most likely, my high school education.
--
Budd Cochran

John 3:16-17, Ephesians 2:8-9
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• posted on March 25, 2006, 1:10 am
Budd Cochran wrote:

Right, you don't have to, but you also can't. The need to shift down has nothing to do with efficiency, and has everything to do with torque multiplication which is an entirely separate topic.

That is where you are wrong. You post one single reputable reference that supports your claim, and I'll concede that I'm wrong and you are right. I don't care what your education level is. I only care that you are claiming physics support for an argument that isn't based on physics. I wouldn't care if you had a Ph.D., I'd still point out the fallacy of your argument.
This is the classic "I lost the logical argument and now need to bail out" statement. No matter what I do you will just keep picking on me. I haven't heard that since grade school.
Matt
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• posted on March 24, 2006, 11:44 am
Budd Cochran wrote:

Less torque, yes, but not less power. Efficiency is related to energy/work/power, not force/torque.

I see the problem now. You are talking about engine efficiency, not gear ratio "efficiency" whatever that is. Keeping the engine in an efficient RPM range is a different discussion altogether, but, again, it doesn't matter how you keep it there (whether OD tranny and numerically higher rear end ratio or 1:1 transmission and a lower numerically rear end ratio. I thought that was your original argument, but you've now changed horses mid-stream. I'm still waiting to see how moving the point of reduction from the rear axle to the transmission makes a fundamental change in efficiency of power transmission.
Matt
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• posted on March 24, 2006, 2:02 pm

Then how come cars running north of me at the Bonneville salt flats run faster with low numerical axles and direct trans ratios than cars with higher numericals and OD ( case in point: Summers Brother's Goldenrod. A still standing land speed record with a car that could not pull in 4th gear, that is OD).

No, I'm talking ratio efficiency, but you keep avoiding / clouding the topic.

You'll never find it. Your head is buried in the sands of overeducation.
--
Budd Cochran

John 3:16-17, Ephesians 2:8-9
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• posted on March 24, 2006, 2:30 pm

Can I get some of that "overeducation"?
Roy

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• posted on March 25, 2006, 12:31 am
Sure, Roy, just follow these two around. It's dripping off them like sweat off a rabbit in hunting season.
--
Budd Cochran

John 3:16-17, Ephesians 2:8-9
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• posted on March 24, 2006, 10:16 pm
Budd Cochran wrote:

Because the OVERALL gear ratio wasn't correct for the power curve of the engine. Ideally, you want the overall ratio to be such that the engine hits its peak horsepower RPM at exactly the same time that the car hits its maximum aerodynamic drag speed. If the car is geared too tall, which this one obviously is, then the aero drag on the car is increasing faster than the power curve of the engine and once they cross, the car will no longer gain speed.
This could easily be fixed by using a numerically higher rear axle ratio. It has nothing to do with the transmission having a ratio less than 1:1, it has everything to do with the OVERALL reduction ratio of the driveline being incorrectly matched to the engine for this particular situation (top speed on the level).

That is because the topic simply doesn't exist. Find us even one reference that talks about a gear ratio efficiency. Just one...

I won't find it because it doesn't exist. And that is the same reason that you can't find it and post a reference to it.
Matt
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• posted on March 25, 2006, 12:39 am

ROTFLMBO!!!! Well, you just stuck your foot in it. According to the Summers Brothers, the car was wind tunnel designed to handle 500 mph, then the engineers determined the engine outputs and ratios needed. And they blew it.

The real life facts call you a liar. Besides, lowering the final drive ratios would have resulted in an unusable first gear due to the slick salt conditions, a slower acceleration and no speed record.

No.
Like I said, and your attitude continues to prove it: it wouldn't matter if I did, mr bigot engineer.
Btw, how about you explain how the marvelous inventions of the past 6000 years came about before your ilk contaminated the Earth?
--
Budd Cochran

John 3:16-17, Ephesians 2:8-9
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• posted on March 25, 2006, 1:19 am
Budd Cochran wrote:

You have yet to show a real life fact. That is why most cars are pushed several hundred yards down the run by another vehicle. Have you ever actually been to Bonneville?

I knew you couldn't, but it has been fun watching you try to weasel out of the hole you've dug.

You mean the attitude that simply is asking you to provide even one fact to back up your claim?
Since you've conceded above that you are wrong and unable to support your claim, my work here is finished. However, Bill is usually more persistent than me so he'll probably entertain your ignorance a little longer. :-)
Matt
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• posted on March 24, 2006, 4:45 am
Budd Cochran wrote:

Methinks you are making the common confusion between power and force (torque). A lever multiplies or divides force (or torque), but it doesn't multiply or divide work or power.
You are the one proposing a violation of the laws of physics. Look up the definition of work and then apply it to both ends of your lever and report back your results.
Matt
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